In a first order reaction, the concentration of reactant decrease from 400 mol `L^(-1)` to 25 mol `L^(-1)` in 200 seconds. The rate constant for the reaction is

Options

A. `1.01386 s^(-1)`
B. `2 xx 10^(-4) s^(-1)`
C. `1.386 xx 10^(-2)s^(-1)`
D. `3.4 xx 10^(-4) s^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
- Initial concentration (A) = 400 mol/L
- Final concentration (A - X) = 25 mol/L
- Time (t) = 200 seconds

### Step 2: Calculate the change in concentration
- Change in concentration (X) = Initial concentration - Final concentration
- X = 400 mol/L - 25 mol/L = 375 mol/L

### Step 3: Use the first-order rate constant formula
For a first-order reaction, the rate constant (k) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{A}{A - X} \right) \]

### Step 4: Substitute the values into the formula
Substituting the values we have:
- \( A = 400 \, \text{mol/L} \)
- \( A - X = 25 \, \text{mol/L} \)
- \( t = 200 \, \text{seconds} \)

So,
\[ k = \frac{2.303}{200} \log \left( \frac{400}{25} \right) \]

### Step 5: Simplify the logarithm
Calculate \( \frac{400}{25} \):
\[ \frac{400}{25} = 16 \]

Now, calculate the logarithm:
\[ \log(16) = 1.204 \]

### Step 6: Substitute back into the equation for k
Now substituting back:
\[ k = \frac{2.303}{200} \times 1.204 \]

### Step 7: Calculate k
Now, perform the calculations:
\[ k = \frac{2.303 \times 1.204}{200} \]
\[ k = \frac{2.773512}{200} \]
\[ k = 0.01386756 \, \text{s}^{-1} \]

### Step 8: Round the answer
Rounding to three significant figures, we get:
\[ k \approx 0.0139 \, \text{s}^{-1} \]

### Final Answer
The rate constant for the reaction is approximately \( k = 1.39 \times 10^{-2} \, \text{s}^{-1} \).

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